Thanks bro it’s really helpful ❤

A solution: void solve() { string s; cin>>s; int n = s.size(); for(int i=0; i<n-2; i++){ cout<<s[i]; } cout<<'i'<<' '; } --------------------------------------------------------------------------------- --------------------------------------------------------------------------------- B solution: void solve() { stack<char>st; string s; cin>>s; int n = s.size(); int cnt = -1; int flag = 0; for(int i=1; i<n; i++){ if(s[i] == s[i-1]){ cnt = i; } } if(cnt == -1) cout<<n; else cout<<1; cout<<' '; // cout<<(n-cn)<<' '; } --------------------------------------------------------------------------------- --------------------------------------------------------------------------------- C solution: void solve() { int n,m; cin>>n>>m; vi a(n); forn(i,0,n) cin>>a[i]; set<int> st; forn(i,0,m){ int x; cin>>x; st.insert(x); } vi b(all(st)); if(b[b.size()-1]-a[n-1] > a[n-1]){ a[n-1] = b[b.size()-1]-a[n-1]; } for(int i=n-2; i>=0; i--){ if(a[i] > a[i+1]){ int val = a[i] + a[i+1]; int ind = lower_bound(all(b), val) - b.begin(); if(ind == b.size()) ind--; if(ind == 0){ if(b[ind] > val){ cout<<"no "; return; }else{ a[i] = b[ind]-a[i]; } }else{ if(b[ind] > val){ a[i] = b[ind-1]-a[i]; }else{ a[i] = b[ind]-a[i]; } } }else if(a[i] < a[i+1]){ int val = a[i] + a[i+1]; int ind = lower_bound(all(b), val) - b.begin(); if(ind == b.size()) ind--; if(ind == 0){ if(b[ind] > val){ // cout<<"no "; // return; }else{ if(a[i] < b[ind]-a[i]){ a[i] = b[ind]-a[i]; } } }else{ if(b[ind] > val){ if(a[i] < b[ind-1]-a[i]){ a[i] = b[ind-1]-a[i]; } }else{ if(a[i] < b[ind]-a[i]){ a[i] = b[ind]-a[i]; } } } } // for(auto j: a) cout<<j<<' '; // cout<<' '; } cout<<"yes "; } --------------------------------------------------------------------------------- --------------------------------------------------------------------------------- D solution: void solve() { int n,m; cin>>n>>m; vector<vi> a(n, vi(m, 0)); for(int i=0; i<n; i++){ for(int j=0; j<m; j++){ cin>>a[i][j]; } } vpp arr; for(int i=0; i<n; i++){ int cnt = 0; int sum = 0; for(int j=0; j<m; j++){ cnt += a[i][j]; // sum += cnt; } arr.pb({cnt, i}); } sort(rall(arr)); vi final; for(auto i: arr){ // cout<<i.ff<<' '<<i.ss<<' '; for(auto k: a[i.ss]){ final.pb(k); } } // for(auto i: final){ // cout<<i<<' '; // }cout<<' '; int ans = 0; int cnt = 0; for(int i=0; i<final.size(); i++){ cnt += final[i]; ans += cnt; } cout<<ans<<' '; } --------------------------------------------------------------------------------- --------------------------------------------------------------------------------- E solution: void solve() { int n,m,k; cin>>n>>m>>k; if(abs(n-m) > k){ cout<<-1<<' '; return; } if(max(n,m) < k){ cout<<"-1 "; return; } int len = n+m; if (n > m){ forn(i,0,k) cout<<0; n -= k; int cnt = 0; forn(i,k,len){ if(cnt == 0 || n == 0) { cout<<1; m--; } else{ cout<<0; n--; } cnt = 1-cnt; } } else { forn(i,0,k) cout<<1; m -= k; int cnt = 0; forn(i,k,len){ if(cnt == 0 || m == 0) { cout<<0; n--; } else{ cout<<1; m--; } cnt = 1-cnt; } } cout<<' '; }

Q galat information de rha bhai

“C:\MinGW\lib\gcc\mingw32\6.3.0\include\c++\mingw32\bits isko paste karne ke bad enter karte hai browser open hota hai

“C:\MinGW\lib\gcc\mingw32\6.3.0\include\c++\mingw32\bits ye kaise kia ,samajh nhi aaya ..paste kar rha hu explorer me kuch nhi ho rha

why did you choose 1 to 100 , not clear

thank you brother

Helpful

Beautiful, thanks brother. Much love. ❤❤

Thanks

Nice approach. Better than tutorial.

Bhai sahi explanation thi Ek suggestion tha, agar script likhkar video shoot karoge tho bolte time kam mistake hoki Baki, looking forward to more such short and crisp explanatory videos

Thanks

B solution: void solve() { int n,m; cin>>n>>m; vector<vi> a(n, vi(m,0)); map<int,set<int>> mp; for(int i=0; i<n; i++){ for(int j=0; j<m; j++){ int x; cin>>x; mp[x].insert(i); } } vector<int> ans; int lastNum = -1; for(int i=0; i<n*m; i++){ int loc = *mp[i].begin(); mp[i].erase(loc); if(ans.size() and ans.back() == loc){ cout<<-1<<' '; return; } ans.pb(loc); if(ans.size() == n){ lastNum = i; break; } } int ind = 0; for(int i=lastNum+1; i<n*m; i++){ int loc = *mp[i].begin(); mp[i].erase(loc); if(loc != ans[ind]){ cout<<-1<<' '; return; } ind = (ind+1)%n; } for(auto i: ans){ cout<<(i+1)<<' '; }cout<<' '; } C solution: void solve() { int n,k; cin>>n>>k; vi a(n); forn(i,0,n) cin>>a[i]; map<int,int> mp; for(auto i: a) mp[i]++; int cnt = 0; for(auto &i: mp){ int sec = k - i.ff; if(mp.count(sec) == 0) continue; if(i.ff == sec){ cnt += mp[i.ff]/2; }else{ int x = min(mp[i.ff], mp[sec]); // cout<<x<<' '; cnt += x; mp[i.ff] -= x; mp[sec] -= x; } } cout<<cnt<<' '; } D solution: void solve() { int n; cin>>n; vi a(n); forn(i,0,n) cin>>a[i]; for(int i=0; i<n-1; i++){ int ind1 = a[i]; int ind2 = a[i+1]; if(ind1 <= ind2){ a[i+1] -= a[i]; a[i] = 0; }else{ cout<<"no"<<' '; return; } } cout<<"yes "; } E solution: class DisjointSet { public: vector<int> rank, parent; DisjointSet(int n) { rank.resize(n + 1, 0); parent.resize(n + 1); for (int i = 0; i <= n; i++) { parent[i] = i; } } int findPar(int node) { if (node == parent[node]) return node; return parent[node] = findPar(parent[node]); } void unionByRank(int u, int v) { int ulp_u = findPar(u); int ulp_v = findPar(v); if (ulp_u == ulp_v) return; if (rank[ulp_u] < rank[ulp_v]) { parent[ulp_u] = ulp_v; } else if (rank[ulp_v] < rank[ulp_u]) { parent[ulp_v] = ulp_u; } else { parent[ulp_v] = ulp_u; rank[ulp_u]++; } } }; void solve() { int n,m1,m2; cin>>n>>m1>>m2; int x = max(m1, m2); DisjointSet *d1 = new DisjointSet(n+1); DisjointSet *d2 = new DisjointSet(n+1); vpp a1, a2; forn(i,0,m1){ int a,b; cin>>a>>b; a1.pb({a,b}); } forn(i,0,m2){ int a,b; cin>>a>>b; a2.pb({a,b}); d2->unionByRank(a, b); } int rem = 0; int add = 0; for(auto &i: a1){ if(d2->findPar(i.ff) == d2->findPar(i.ss)){ d1->unionByRank(i.ff, i.ss); continue; }else{ i.ff = 0; i.ss = 0; rem++; } } for(auto &i: a2){ if(d1->findPar(i.ff) == d1->findPar(i.ss)) continue; d1->unionByRank(i.ff, i.ss); add++; } cout<<add+rem<<' '; }

Thanks bhai

Thanks man :)

Thank you so much!

B solution: void solve() { int n; cin>>n; vi a(n); forn(i,0,n) cin>>a[i]; vi b(n); forn(i,0,n) cin>>b[i]; int maxInc = 0; int ind = 0; forn(i,0,n){ if(a[i] < b[i] && abs(a[i]-b[i]) >= maxInc){ ind = i; maxInc = abs(a[i]-b[i]); } } forn(i,0,n){ if(i!=ind && a[i]-maxInc < b[i]){ cout<<"no "; return; } } cout<<"yes "; } C solution: void solve() { int n,m; cin >> n >>m; string s; cin >> s; vector<vector<int>> v(n,vector<int>(m)); for(int i = 0;i<n;i++) { for(int j =0;j<m;j++) { int num; cin >> num; v[i][j] = num; } } int curi = 0; int curj = 0; for(int i = 0;i<s.size();i++) { int sum = 0; if(s[i]=='D') { for(int j = 0;j<m;j++) { sum+=v[curi][j]; } v[curi][curj] = -sum; curi++; } else{ for(int k = 0;k<n;k++) { sum+=v[k][curj]; } v[curi][curj] = -sum; curj++; } } int sum2 = 0; for(int j = 0;j<m;j++) { sum2+=v[n-1][j]; } v[n-1][m-1] = -sum2; for(int i = 0;i<n;i++) { for(int j = 0;j<m;j++) { cout << v[i][j] << ' '; } cout << ' '; } cout << ' '; }

Nice explanation

Most underrated channel on KZbin Seriously!!! Salute to you bro have been following you since from long time

thanks man

👌🏻

I have a question guys , I am really confused whether to learn dsa using java or python , python feels simple and easy but java is verbose. Which will be better or maybe useful

its actually cool

Worked for me if the repo is PUBLIC not private

Really cool man Keep making such content

great video anything similar for codeforces, codechef or gfg??

I've been syncing my LeetCode progress manually to GitHub, but interesting that there is an extension to do this!

cool

Does it also copies link of the question which one solved from leetcode to github?

Having windows error 2 file specified not found

awesome bro ..... maja a geya ... really ..🥰💓

hello bhai , tum shi mein bhot madad kar rhe ho, but pls kya tum javascript by jonas schmedtmann ya phir angela yu ka latwst course de sakte ho web development ka

I am the 1000th subscriber. Hope to see more of your content. Keep going buddy !!!

Bc sahi se bata de c batana ho to

How did the formula for C came ??? Dont call a "reading out" of solution an editorial.

Nice explanation. Thanks bhai!

6:46 it should be 0110.

thanks a lot bro well explained!!

Great explanation brother

brother youre doing really great . how can we perform good in contests

nice work brother

please see this image for c's case with k==2 drive.google.com/file/d/1YTUALw1qMOlqAcJeX4855HkIq0pgIqXU/view?usp=sharing B solution: void solve() { int n; cin>>n; map<int,int> mp; vpp a(n); set<int> st; forn(i,0,n){ cin>>a[i].ff; cin>>a[i].ss; if(a[i].ff == a[i].ss){ st.insert(a[i].ff); mp[a[i].ff]++; } } vi temp (all(st)); vector<int> ans(n); for(int i=0; i<n; i++){ if(a[i].ff == a[i].ss){ if(mp[a[i].ff] > 1){ ans[i] = 0; }else{ ans[i] = 1; } }else{ int ind1 = lower_bound(all(temp), a[i].ff) - temp.begin(); int ind2 = lower_bound(all(temp), a[i].ss) - temp.begin(); if(ind1 != temp.size() and ind2 != temp.size() and temp[ind1] == a[i].ff and temp[ind2] == a[i].ss and ind2-ind1+1 == a[i].ss - a[i].ff + 1 ){ ans[i] = 0; }else{ ans[i] = 1; } } } for(auto i: ans){ cout<<i; } cout<<' '; } C solution: pair<int,int> f(int l, int r, int k) { if (r-l+1 < k) { return {0, 0}; } int mid = l+(r-l)/2; if ((r-l+1)%2){ pair<int,int> val = f(l, mid - 1, k); int sum = mid + 2 * val.ff + (mid * val.ss); int segNum = 2 * val.ss + 1; return {sum, segNum}; }else{ pair<int, int> val = f(l, mid, k); int sum = 2 * val.ff + mid * val.ss; int segNum = 2 * val.ss; return {sum, segNum}; } } void solve() { int n,k; cin>>n>>k; cout<<f(1, n, k).ff<<' '; } D solution: const int maxn = 1e6; int a[maxn] , b[maxn]; int powmod(int x , int y){ if(y == 0) return 1; int m = y / 2; int res = powmod(x , m); res = (res * res) % mod; if(y%2 == 1) res = (res * x) % mod; return res; } void solve() { int n , q; cin >> n >> q; vector<int> v1 , v2; for(int i = 0 ; i < n ; i ++ ){ cin >> a[i]; v1.pb(a[i]); } for(int i = 0 ; i < n ; i ++ ){ cin >> b[i]; v2.pb(b[i]); } sort(v1.begin() , v1.end()); sort(v2.begin() , v2.end()); map<int,int> lastA , lastB , cntA , cntB; int res = 1; for(int i = 0 ; i < n ; i ++ ){ lastA[v1[i]] = i; lastB[v2[i]] = i; cntA[v1[i]] ++ ; cntB[v2[i]] ++ ; res = res * min(v1[i] , v2[i]); res = res % mod; } vector<int> ans; ans.pb(res); while(q--){ int type, ind; cin>>type>>ind; ind-- ; if(type == 1){ int x = a[ind]; int i = lastA[x]; cntA[x]--; if(cntA[x] == 0) lastA[x] = -1; else lastA[x]--; res = (res * powmod(min(v1[i] , v2[i]) , mod-2)) % mod; v1[i]++; a[ind]++; cntA[v1[i]]++; if(cntA[v1[i]] == 1) lastA[a[ind]] = i; res = res*min(v1[i] , v2[i]); res %= mod; } else{ int x = b[ind]; int i = lastB[x]; cntB[x] -- ; if(cntB[x] == 0) lastB[x] = 0; else lastB[x] -- ; res = res * powmod(min(v1[i] , v2[i]) , mod - 2) % mod; v2[i]++; b[ind]++; cntB[v2[i]]++; if(cntB[v2[i]] == 1) lastB[b[ind]] = i; res = res *min(v1[i] , v2[i]); res %= mod; } ans.pb(res); } for(auto x : ans) cout << x << " "; cout<<' '; }

Explanation why for C limit is 2*n ? (only see if you understood all other part and point of the question ) See, its actually simple when you realize. There we are given A with length n right ? What is the best score we can get out of A by performing PREFIX ADD operations ? lets say X , now what is the maximum value of X ? obviously "n" right ? because there are "n" elements in "a" , so it makes sense. to score "n" without optimizing , if we blindly go NO BRAINER approach like below seq ADD , SCORE , ADD, SCORE, ADD ..... like this, it will take like 2*n operations (n add n score) right to reach "n" score ? that's why we can't afford more than 2*n operations on figuring out the most optimal solution. if 2*n operations already done, that best we could have achieved was n , and that same can be achieved by the no brainer approach I mentioned above so we won't calculate further.

Situation Task ,action,result (STAR) -witht he resources avaiable in online, posting thse videos help us prepare rigorously and ace the interviews excellently..

Brilliant techie!!! pivoting to diffent skillsets - java spring boot data science, android deelopent- woman security app, placement app, is your next career- cryptography engineer or algorithm designer

thanks bhai it works

From which IIT, she was passed out

bhai teri cf profiile?

Can you show your profile link